One for the mathematicians

For my bowls club I run a monthly 100 Club draw with three cash prizes. The numbers being randomly drawn from a bag..

Yesterday evening at the annual club opening social a draw was made;

Against probability two members drew their own numbers out of the bag. I wonder what the odds on this happening are?

There are 79 numbers in "play" not all consecutive, the numbers missing from 1 to 100 are, should it make any difference to the calculation:6,10,15,22,25,26,27,30,33,35,36,38,40,41,44,51,58,66,67,70,71,72,74,75,76,77,80,90,93,and 94.

Nine numbers included over 100 are 104,105,106,107,110,113,115,120 and 121.

Over the years there used to be a lot more players hence the numbers over 100, those remaining preferred to keep their numbers.

Experience; is something you gain, just after you needed it most.

When faced with two choices, simply toss a coin. It works not because it settles the question for you. But because in that brief moment while the coin is in the air. You suddenly know what you are hoping for.

Re: One for the mathematicians

Ah! the bit I forgot to add is: there maybe 79 numbers in play but, some members have more than one number.

So there are 73 actual players.

Experience; is something you gain, just after you needed it most.

When faced with two choices, simply toss a coin. It works not because it settles the question for you. But because in that brief moment while the coin is in the air. You suddenly know what you are hoping for.

Re: One for the mathematicians

Do the members with more than one ticket all have two or do some have more? I think that matters. Assuming that the three draws are made by different members... A ticket is drawn by an owner with two tickets. That means that their remaining ticket cannot be drawn by its owner. However, if a ticket is drawn by an owner with five tickets then there are four remaining that cannot be drawn by their owner.

I'm afraid I've forgotten most of the statistics that I learned at school and this problem is too complicated for me to solve.

BBC Radio 4's More Or Less may be of interest to you if you want a better understanding of statistics. It also occasionally answers listeners' questions.

Re: One for the mathematicians

No alanf, both players have only one number in their own name. It happens that one of them has numbers for other family members in their own names amounting to six numbers in all, but only one in this winners name.

Experience; is something you gain, just after you needed it most.

When faced with two choices, simply toss a coin. It works not because it settles the question for you. But because in that brief moment while the coin is in the air. You suddenly know what you are hoping for.

Re: One for the mathematicians

Y'know Browni you make that seem rediculously simple. I would guess it has to be right though. Thank you.

Experience; is something you gain, just after you needed it most.

When faced with two choices, simply toss a coin. It works not because it settles the question for you. But because in that brief moment while the coin is in the air. You suddenly know what you are hoping for.

Re: One for the mathematicians

If there are 79 numbers in play the odds that 1 player will pick his own number are 1 in 79.

The odds for a second player picking his own number are 1 in 78 as 1 ball has already been picked.

Multiplying them together gives odds of 1 in 6,162 of this happening together.

Very authoritative, but incorrect! You have to divide results by the factorial of 2 i.e 2x1. were there three numbers from 79 you would need to divide the product of 79x78x77 by factorial of 3 i.e. 3x2x1.

Someone didn't read the link I supplied!

No one has to agree with my opinion, but in the time I have left a miracle would be nice.

Re: One for the mathematicians

@Luzern Having done the football pools before the Lotto started I'm familiar with the calculation in your post, but that is for drawing any two from 79. Here we are talking about two separate draws of one from 79 followed by one from 78.

Now here's something I've wondered about: On Deal or No Deal 22 boxes are each selected at random by contestants at the beginning of the game. One contestant out of the 22 is then selected to play the game. Is the chance of that player having the top prize in his box one in 22 or one in 22x22?

Re: One for the mathematicians

Now here's something I've wondered about: On Deal or No Deal 22 boxes are each selected at random by contestants at the beginning of the game. One contestant out of the 22 is then selected to play the game. Is the chance of that player having the top prize in his box one in 22 or one in 22x22?

If the contestant has already been chosen then there is 1 in 22 chance that they have the top prize.

If there are 22 potential contestants then there is a 1 in 22 chance that they will be chosen and then a 1 in 22 chance that they have the top prize so then the odds are 1 in 22x22.